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Let  tan(15°) = tan(45°-30°)

We know that tan(A - B) = (tanA - tanB) /(1 + tan A tan B)

⇒tan(45°-30°) = (tan45°- tan30°)/(1+tan45°tan30°)

= {1- (1/√3)} / {1+(1/√3)}

∴ tan15° = (√3 - 1) / (√3 + 1)

Given:

Sinx = 3/5 

Formula Used:

sinθ = perpendicular/hypotenuse 

cosθ = Base/hypotenuse

cotθ = cosθ/sinθ  

Calculation:

cotx × secx = (cosx/sinx) × 1/cosx = 1/sinx 

⇒ cotx × secx = 1/(3/5) = 5/3

∴ The value of  ​\(\cot x.\sec x\) is 5/3

Formula used:

sin 2A = 2tanA/(1 + tan²A)

cos 2A = (1 - tan2A)/(1 + tan2A)

Calculation:

[(1 + tan2A)/2tan A] × [(1 - tan2A)/(1 + tan2A)]

⇒ (1/sin 2A) × cos 2A

⇒ cos 2A/sin 2A

⇒ cot 2A

Formula Used:

Sin(A +B) = SinACosB + CosASinB

Cos(A – B) = CosACosB + SinASinB

Calculation:

⇒ Sin(60 + θ) = Sin60×Cosθ + Cos60×Sinθ

⇒ Sin(60 + θ) = (√3/2)Cosθ + (1/2)Sinθ       ----(1)

⇒ Cos(30 – θ) = Cos30×Cosθ + Sin30×Sinθ

⇒ Cos(30 – θ) = (√3/2)Cosθ + (1/2)Sinθ      ----(2)

⇒ Now the value of Sin(60 + θ) – Cos(30 – θ) from (1) & (2)

⇒ (√3/2)Cosθ + (1/2)Sinθ – (√3/2)Cosθ - (1/2)Sinθ = 0

∴ The answer is 0 

Short Trick:

Put θ = 0°, we get 

⇒ Sin60 - Cos30 = 0

∴ From both we get 0

Given:

3sinθ + 4cosθ = 5 we have to find the value of cotθ

Concept Used:

Sin²θ + cos²θ = 1

Cotθ = cosθ/sinθ

Calculation:

3sinθ + 4cosθ = 5

⇒ (3/5)sinθ + (4/5)cosθ = 1      ----(1)

We know that, Sin²θ + cos²θ = 1      ----(2)

Comparing (1) and (2) get,

Sinθ = 3/5 and cosθ = 4/5

Cotθ = cosθ/sinθ

⇒ cotθ = (4/5)/(3/5)

⇒ cotθ = 4/3

∴ Value of cotθ is 4/3

Concept used :

\(\pi \;radian = 180^\circ \)

sin 30° = 1/2

cosec 30° = 2

Solution :

\(\\sin \left( {\frac{\pi }{6}} \right) + cosec\left( {\frac{\pi }{6}} \right)\)

⇒ sin 30° + cosec 30°

⇒ (1/2) + 2

⇒ 5/2

∴ The value of \(\sin \left( {\frac{\pi }{6}} \right) + cosec\left( {\frac{\pi }{6}} \right)\) is 5/2.

Given:

 Our given expression is \(\frac{{cos\theta \; + \;sin\theta }}{{\sqrt {1\; + \;sin2\theta } }}\)

Formula used:

sin2θ = 2sinθ cosθ

sin²θ + cos²θ = 1

(a + b)² = a² + b² + 2ab

Calculation:

Our given expression is \(\frac{{cos\theta \; + \;sin\theta }}{{\sqrt {1\; + \;sin2\theta } }}\) 

⇒ (cosθ + sinθ)/√(sin²θ + cos²θ + 2sinθ cosθ)

⇒ (cosθ + sinθ)/√(sinθ + cosθ)²

⇒ (cosθ + sinθ)/(sinθ + cosθ)

⇒ 1

∴ The value of the given expression is 1

Given:

\( \frac{{\cos x}}{{1 + \sin x}} + \frac{{1 + \sin x}}{{\cos x}}\)

Concept Used:

sin2x + cos2x = 1

Calculation:

\( \frac{{\cos x}}{{1 + \sin x}} + \frac{{1 + \sin x}}{{\cos x}}\)

⇒ {cos²x + (1 + sinx)²}/{(1 + sinx) × cosx}

⇒ (cos²x + 1 + sin²x + 2sinx)/{(1 + sinx) × cosx}

⇒ (1 + 1 + 2sinx)/{(1 + sinx) × cosx}              [∵ sin²x + cos²x = 1]

⇒ 2(1 + sinx)/{(1 + sinx) × cosx}

⇒ 2/cosx

⇒ 2secx

∴ \( \frac{{\cos x}}{{1 + \sin x}} + \frac{{1 + \sin x}}{{\cos x}}\) = 2secx

Concept:

tan 45° = 1

Calculation:

sin θ - cos θ = 0 

⇒ sin θ = cos θ 

⇒ \(\rm \frac{\sin θ }{\cos θ }=1\)

⇒ tan θ  = 1 

So, θ = 45° 

Hence, option (2) is correct. 

Concept:

\(\rm sin ^2θ +cos^2θ =1\)

Calculation:

Here, sin θ = 3/5, 

We know, \(\rm sin ^2θ +cos^2θ =1\)

\(\rm \Rightarrow cos^2θ =1 -(\frac 3 5)^2\\ =\frac{25-9}{25}\)

= 16/25

∴ cos θ  = 4/5

Hence, option (3) is correct.

Concept:

sin 2θ = 2 sin θ cos θ 

Calculation: 

We know, sin 2A = 2 sin A cos A 

Given: sin 2A = 2 sin A

⇒ 2 sin A cos A  = 2 sin A

⇒ 2 sin A cos A - 2 sin A = 0

⇒ 2 sin A(cos A - 1) = 0

So, cos A = 1,when A = 0° 

And 2 sin A = 0 when A= 0° 

∴ If sin 2A = 2 sin A then  A = 0° 

Hence, option (1) is correct.

Concept:

\(\rm 1+\tan ^2θ =sec^2θ\)

Calculation:

Given: tan θ = √5

We know that, \(\rm 1+\tan ^2θ =sec^2θ\)

\(\rm sec^2θ = 1 + (√ 5)^2=1+5\)

= 6

∴ sec θ = √6

Hence, option (3) is correct. 

tan x = 1 , but as tan (π/4) = 1 

However, as period of tan function is π and the function repeats after every π , the general solution for tan x = 1 = tan (π/4) is 

x = nπ + π/4 , where n is an integer.

Use trig table of special arcs: 

When tan x = - 1

x = 3π/4

General answers: 

x = 3π/4 + kπ

Given:

(cos θ + sin θ)/(cos θ - sin θ) = 8

Formula:

If x/y = a/b, then

Dividendo and Componendo rule

(x + y)/(x - y) = (a + b)/(a - b)

Calculation:

(cos θ + sin θ)/(cos θ - sin θ) = 8/1

Dividendo and Componendo rule

⇒ cos θ/sin θ = (8 + 1)/(8 - 1)

⇒ cot θ = 9/7

Concept:

The identities of trigonometry are:

• \(\rm \tan a={\sin a\over \cos a}\)
• \(\rm \cot a ={\cos a\over\sin a} = {1\over \tan a}\)
• \(\rm \tan (a+b)={\tan a +\tan b\over{1-\tan a \tan b}}\)
• \(\rm \tan (a-b)={\tan a -\tan b\over{1+\tan a \tan b}}\)

Calculation:

Given 

cot B - cot A = y

⇒ \(\rm {1\over\tan B}-{1\over\tan A} =y\)

⇒ \(\rm {\tan A - \tan B\over\tan A\tan B} =y\)

Given tan A - tan B = x

⇒ \(\rm {x\over\tan A\tan B} =y\)

⇒ \(\boldsymbol{\rm \tan A\tan B ={x\over y}}\)

Now, \(\rm \tan (A-B)={\tan A -\tan B\over{1+\tan A \tan B}}\)

⇒ tan (A - B) = \(\rm {x\over{1+{x\over y}}}\) 

⇒ tan (A - B) = \(\rm {xy\over{x+y}}\)

cot (A - B) = \(\rm 1\over\tan (A - B)\) 

⇒ cot (A - B) = \(\rm x+y\over xy\) 

⇒ cot (A - B) = \(\boldsymbol{\rm {1\over y} + {1\over x}}\) 

Given:

Sin D = 3/5, we have to find the value of (sin D + cos D)2

Concept Used:

Sinθ = Height/Hypotenuse

Cosθ = Base/Hypotenuse

In a Right angle triangle, Height² + Base² = Hypotenuse

Calculation:

 Sin D = 3/5 = Height/Hypotenuse

⇒ Height = 3k and Hypotenuse = 5k     [Where k is a constant]

We know,

Height2 + Base2 = Hypotenuse

⇒ (3x)² + Base² = (5x)²

⇒ Base² = 25x² - 9x²

⇒ Base2 = 16x²

⇒ Base = 4x

Cos D = 4x/5x

⇒ Cos D = 4/5

(sin D + cos D)2

⇒ (3/5 + 4/5)²

⇒ (7/5)²

⇒ 49/25

∴ The required value of (sin D + cos D)² is 49/25.

Given:

degree = 72° 30'

Concept Used:

Radian measure = (π/180) × degree measure

1 degree = 60 minutes

Calculation:

Value of 72° 30' = (72 + 30/60)  

Value = (72 + 1/2) = (145/2)° 

Substituting the values in the formula 

Radian measure of 72° 30' = (145/2) × (π/180)

Radian measure of 72° 30' =  29π/72

∴ The radian measure of 72° 30' is 29π/72.

Given:

4θ is an acute angle,

And cot 4θ = tan (θ - 5°).

Formula used:

tan (90° - θ) = cot θ 

Calculation:

cot 4θ = tan (θ -5°)

⇒ tan (90° - 4θ) = tan (θ -5°)

⇒ 90° - 4θ = θ -5° 

⇒ 90° + 5° = θ + 4θ 

⇒ 95° = 5θ 

⇒ θ = 19° 

∴ the value of θ is 19°.

Given:

It is given that 2 sin θ = x and x sec θ = 2

Formula Used:

Basic concept of trigonometric ratio and identities

We know that

tanθ = sinθ/cosθ

secθ = 1/cosθ

Calculation:

2sinθ = x

∴ sinθ = x/2

And x secθ = 2

∴ Secθ = 2/x

⇒ cosθ = x/2

Now, we have to find the value of 2tanθ + 1 = 2(sinθ/cosθ) + 1

Put the value of sin θ and cos θ in the given expression

∴ 2[(x/2)/(x/2)] + 1 = 2 + 1 = 3

Hence, option (3) is correct

Formula Used:

Sin(A ± B) = SinACosB ± SinBCosA

Cos(A + B) = CosACosB - SinBsinA

Cos(A - B) = CosACosB + SinBsinA

Calculation:

We have to calculate the value of Sin 75°

∴ Sin 75° = Sin (45 + 30)° = Sin45°Cos30° + Sin30°Cos45°

⇒ Sin 75° = 1/√2 × √3/2 + 1/2 × 1/√2 = (√3 + 1)/2√2

So, statement 1 is correct

Now the value of Cos 75°

∴ Cos 75° = cos(45 - 30)° = Cos45°Cos30° - Sin45°sin30°

⇒ Cos 75° = 1/√2 × √3/2 - 1/2 × 1/√2 = (√3 - 1)/2√2

So, statement 2 is correct

Hence, option (3) is correct

Given

2sec2θ + tan2θ = 17 

Concept

Sec²θ - tan²θ = 1

Sec²θ = 1 + tan²θ 

Cotθ = (1/tanθ)

Calculation

 We have 2sec2θ + tan2θ = 17

⇒ 2(1 + tan²θ ) + tan²θ = 17

⇒ 2 + 2tan²θ + tan²θ = 17

⇒ 3tan²θ = 15

⇒ tan²θ = (15/3)

⇒ tan²θ = 5 

⇒ tanθ = √5 

⇒ cotθ = (1/√5)

∴ Cotθ = (1/√5)

Given:

It is given that tan x + tan y = 2

Formula Used:

Basic concept of trigonometric ratio and identities

We know that

tan45° = 1 and cot 45° = 1

Calculation:

Let x = 45° and y = 45°

∴ tan 45° + tan 45° = 1 + 1 = 2

Now, cot x – cot y = cot 45° – cot 45° = 1 – 1 = 0

Hence, option (2) is correct

Given:

It is given that sin (π/2 - θ/3) = √3/2

Formula Used:

Basic concept of trigonometric ratio and identities

We know that

sin(90 -θ) = cosθ

⇒ cos30° = √3/2

Calculation:

We know that sin (π /2 - θ) = cos θ

∴ sin(π/2 - θ/3) = √3/2

⇒ cos(θ/3) = cos 30 °

⇒ θ/3 = 30 ° = θ = 90 °

Now, value of tan θ = tan 90 ° = Not define

Hence, option (4) is correct

Concept

Sin(90°- A) = Cos A

Cos (90°- A) = Sin A

Calculation

Sin(90°– 70°) Cos 70°+ Cos(90° – 70°) Sin 70°

⇒ Cos 70°Cos 70°+ Sin 70°Sin 70°

⇒ Cos²70 °+ Sin²70°

Given:

sin²38^° – cos²52^° 

Concept Used:

sin²θ + cos²θ = 1

sin(90^° – θ ) = cosθ 

Calculation:

sin²38^° – cos²52^° = sin²38^° – sin²(90^° – 52^°)

⇒ sin²38^° – sin²38^°

⇒ 0

⇒ sin²38^° – cos²52^° = 0

∴ sin²38^° – cos²52^° = 0

Given:

sec B + tan B = r

Formula Used:

sec²θ - tan²θ = 1

a² - b² = (a + b) (a - b)

Calculation:

sec² B - tan² B = 1 

⇒ (sec B + tan B) (sec B - tan B) = 1

⇒ (r) (sec B - tan B) = 1 

⇒ sec B - tan B = 1/r

Concept:

tan(A - B) = \(\rm \frac{tanA-tan B}{1+tan AtanB}\)

Calculation:

We know, tan(A - B) = \(\rm \frac{tanA-tan B}{1+tan AtanB}\)

∴cot(A - B) = \(\rm \frac{1+tan AtanB}{tanA-tan B}\)

\(=\rm \frac{1+\frac{3}{4}\times \frac{-12}{5}}{\frac 3 4 + \frac{12}{5}}\)

\(=\rm \frac{1-\frac{36}{20}}{\frac {63}{20}}\)

\(=\rm \frac{-16}{20}\times \frac {20}{63}=\frac{-16}{63}\)

Given

cot A = tan(2A - 45°)

Concept

If cot A = tan(90° - A)

A = 90° - A

Calculation

cot A = tan(2A - 45°)

⇒ A = 2A - 45° 

⇒ A = 45° 

⇒ tan A = tan 45° 

Given:

\(\frac{{{{\sin }^2}{}52^\circ + 2 + {{\sin }^2}{{ }}38^\circ }}{{4{}{{\cos }^2}43^\circ - 5 + 4{{\cos }^2}{{}}47^\circ }}\)

Identity used:

Sin θ = cos(90 - θ)

Sin² θ + cos² θ = 1

Calculation:

\(\frac{{{{\sin }^2}{}52^\circ + 2 + {{\sin }^2}{{ }}38^\circ }}{{4{}{{\cos }^2}43^\circ - 5 + 4{{\cos }^2}{{}}47^\circ }}\)

using Sin θ = cos(90 - θ)

⇒ \(\frac{{si{n^2}52^\circ \; + \;2\; + \;si{n^2}\left( {90 - 52} \right)^\circ }}{{4co{s^2}43^\circ - 5\; + \;co{s^2}\left( {90\; - 43} \right)^\circ }}\)

⇒ \(\frac{{si{n^2}52^\circ \; + \;2\; + \;co{s^2}52^\circ }}{{4co{s^2}43^\circ - 5\; + \;si{n^2}43^\circ }}\)

Applying Sin2 θ + cos2 θ = 1

⇒ (1 + 2)/(4 - 5) ⇒ -3

∴ The value of \(\frac{{{{\sin }^2}{}52^\circ + 2 + {{\sin }^2}{{ }}38^\circ }}{{4{}{{\cos }^2}43^\circ - 5 + 4{{\cos }^2}{{}}47^\circ }}\) is -3.

Given:

\(\sin A = \dfrac{15}{17}\) and \(\sin B = \dfrac{7}{25}\)

Formula used:

sin(A - B) = sinAcosB - cosAsinB

Calculation:

\(\sin A = \dfrac{15}{17}\), then \(\cos A = \dfrac{8}{17}\)

\(\sin B = \dfrac{7}{25}\), then \(\cos B = \dfrac{24}{25}\)

sin(A - B) = sinAcosB - cosAsinB

⇒ (15/17) × (24/25) - (8/17) × (7/25)

⇒ (72/85) - (56/425)

⇒ (360 - 56)/425

⇒ 304/425

∴ The value of sin(A - B) is 304/425.

It is known that the values of cosec x repeat after an interval of 2π or 360°.

cosec(−1410^∘)=−cosec1410^∘

⇒−cosec(4×360^∘−30^∘)

⇒ - cosec30^∘

Since cosec(2nπ−θ)=−cosecθ

cosec30^∘⇒2

Hence cosec(−1410^∘)=2